Question

Решите на 4 плз) для правдободобности

Answer 1

Довольно простое задание.

Answer 2

Адказ:

4.

1)

$3 \sqrt{7} = \sqrt{3 \times 3 \times 7} = \sqrt{21}$

$4 \sqrt{6} = \sqrt{4 \times 4 \times 6} = \sqrt{96}$

$\sqrt{21} < \sqrt{96}$

2)

$5 \sqrt{ \frac{7}{5} } = \sqrt{5 \times 5 \times \frac{7}{5} } = \sqrt{5 \times 7} = \sqrt{35}$

$\frac{1}{2} \sqrt{140} = \sqrt{ \frac{1}{2} \times \frac{1}{2} \times 140 } = \sqrt{ \frac{1}{4} \times \frac{140}{1} } = \sqrt{ \frac{140}{4} } = \sqrt{35}$

$\sqrt{35} = \sqrt{35}$

5.

1)

$\frac{12}{ 7\sqrt{3} } = \frac{12}{ 7\sqrt{3} } \times \frac{7 \sqrt{3} }{ 7\sqrt{3} } = \frac{12 \times 7 \times \sqrt{3} }{7 \times 7 \times \sqrt{3} \times \sqrt{3} } = \frac{84 \sqrt{3} }{49 \times 3} = \frac{84 \sqrt{3} }{147}$

2)

$\frac{18}{ \sqrt{7} + 1 } = \frac{18}{ \sqrt{7} + 1 } \times \frac{ \sqrt{7} - 1}{ \sqrt{7} - 1} = \frac{18 \sqrt{7} - 18 }{7 - 1} = \frac{18 \sqrt{7} - 18}{6}$

Поспехаў!